As $\rm x$ varies from $\rm −1\ to \ +3$, which one of the following describes the behaviour of the function $\rm f(x) = x^3 – 3x^2 + 1$?

Question

As $\rm x$ varies from $\rm &minus;1\ to \ +3$, which one of the following describes the behaviour of the function $\rm f(x) = x^3 &ndash; 3x^2 + 1$?

Prepp Team · Accepted Answer

Function Behavior Analysis

To understand the behavior of the function $ \rm f(x) = x^3 – 3x^2 + 1 $ as $ \rm x $ varies from $ \rm −1 $ to $ \rm +3 $, we need to analyze its first derivative. The first derivative, $ \rm f'(x) $, provides crucial information about where the function is increasing or decreasing.

Derivative Calculation

First, let's find the derivative of the given function $ \rm f(x) $.

The given function is: $ \rm f(x) = x^3 – 3x^2 + 1 $

To find the first derivative $ \rm f'(x) $, we differentiate $ \rm f(x) $ with respect to $ \rm x $:

$ \rm f'(x) = \frac{d}{dx}(x^3 – 3x^2 + 1) $

$ \rm f'(x) = 3x^{3-1} – 3(2x^{2-1}) + 0 $

$ \rm f'(x) = 3x^2 – 6x $

Critical Points Identification

Next, we identify the critical points of the function. Critical points are the values of $ \rm x $ where the first derivative $ \rm f'(x) $ is equal to zero or undefined. These points often indicate where the function changes its behavior from increasing to decreasing or vice versa.

Set $ \rm f'(x) = 0 $:

$ \rm 3x^2 – 6x = 0 $

Factor out the common term $ \rm 3x $:

$ \rm 3x(x – 2) = 0 $

This equation yields two possible values for $ \rm x $:

$ \rm 3x = 0 \implies x = 0 $
$ \rm x – 2 = 0 \implies x = 2 $

Both of these critical points, $ \rm x = 0 $ and $ \rm x = 2 $, fall within the specified interval for $ \rm x $, which is $ \rm [-1, 3] $.

Interval Behavior Analysis

Now, we analyze the sign of $ \rm f'(x) $ in the sub-intervals created by these critical points within the given range $ \rm [-1, 3] $. The relevant intervals are $ \rm (-1, 0) $, $ \rm (0, 2) $, and $ \rm (2, 3) $.

Interval	Test Value ($ \rm x $)	Expression for $ \rm f'(x) = 3x(x – 2) $	Sign of $ \rm f'(x) $	Behavior of $ \rm f(x) $
$ \rm (-1, 0) $	$ \rm x = -0.5 $	$ \rm 3(-0.5)(-0.5 – 2) = 3(-0.5)(-2.5) = 3(1.25) = 3.75 $	$ \rm f'(x) > 0 $	Increasing
$ \rm (0, 2) $	$ \rm x = 1 $	$ \rm 3(1)(1 – 2) = 3(1)(-1) = -3 $	$ \rm f'(x) < 0 $	Decreasing
$ \rm (2, 3) $	$ \rm x = 2.5 $	$ \rm 3(2.5)(2.5 – 2) = 3(2.5)(0.5) = 3(1.25) = 3.75 $	$ \rm f'(x) > 0 $	Increasing

Function Behavior Summary

Based on our analysis of the sign of $ \rm f'(x) $ in each interval, we can summarize the behavior of the function $ \rm f(x) $:

As $ \rm x $ goes from $ \rm -1 $ up to $ \rm 0 $, the function $ \rm f(x) $ is increasing.
As $ \rm x $ goes from $ \rm 0 $ up to $ \rm 2 $, the function $ \rm f(x) $ is decreasing.
As $ \rm x $ goes from $ \rm 2 $ up to $ \rm 3 $, the function $ \rm f(x) $ is increasing again.

Conclusion on Function Behavior

Therefore, as $ \rm x $ varies from $ \rm -1 $ to $ \rm +3 $, the behavior of the function $ \rm f(x) = x^3 – 3x^2 + 1 $ can be described as first increasing, then decreasing, and finally increasing again. This comprehensive analysis details the exact behavior of the function $ \rm f(x) $ across the specified range.

Prepp Team · Answer

$\rm f(x)$ increases monotonically.

Prepp Team · Answer

$\rm f(x)$ decreases, then increases and decreases again.

Prepp Team · Answer

$\rm f(x)$ increases and then decreases.

Interval	Test Value (\( \rm x \))	Expression for \( \rm f'(x) = 3x(x – 2) \)	Sign of \( \rm f'(x) \)	Behavior of \( \rm f(x) \)
\( \rm (-1, 0) \)	\( \rm x = -0.5 \)	\( \rm 3(-0.5)(-0.5 – 2) = 3(-0.5)(-2.5) = 3(1.25) = 3.75 \)	\( \rm f'(x) > 0 \)	Increasing
\( \rm (0, 2) \)	\( \rm x = 1 \)	\( \rm 3(1)(1 – 2) = 3(1)(-1) = -3 \)	\( \rm f'(x) < 0 \)	Decreasing
\( \rm (2, 3) \)	\( \rm x = 2.5 \)	\( \rm 3(2.5)(2.5 – 2) = 3(2.5)(0.5) = 3(1.25) = 3.75 \)	\( \rm f'(x) > 0 \)	Increasing